Solve and check result: \(2y+\frac{5}{3}=\frac{26}{3}-y\)
\(2y+\frac{5}{3}=\frac{26}{3}-y\)
Transposing \(y\) to L.H.S and \(\frac{5}{3}\) to R.H.S, we obtain
\(2y+y=\frac{26}{3}-\frac{5}{3}\)
\(3y\) = \(\frac{21}{3}\) = \(7\)
Dividing both sides by \(3\), we obtain
\(y\) = \(\frac{7}{3}\)
L.H.S = \(2y+\frac{5}{3}\)
=\(2×\frac{7}{3}+\frac{5}{3}\)
=\(\frac{14}{3}+\frac{5}{3}\)
=\(\frac{19}{3}\)
R.H.S = \(\frac{26}{3}-y\)
=\(\frac{26}{3}-\frac{7}{3}\)
=\(\frac{19}{3}\)
L.H.S. = R.H.S.
Hence, the result obtained above is correct.